Can't I always create regular $n$ sided polygon for any given value of $n$? Following is the first paragraph from the section "Construction of regular polygon" from Hardy and Wright. > We conclude this chapter by a short excursus on one of the famous problems of elementary geometry, that of the construction of a regular polygon of $n$ sides, or of an angle $\alpha = \frac{2\pi}{n}$. Suppose that $gcd(n_1,n_2) = 1$ and that the problem is soluble for $n = n_1$ and for $n = n_2$. There are integers $r_1$ and $r_2$ such that $$r_1n_1,+r_2n_2 = 1 \;...(1)$$ or $$r_1\alpha_2+r_2\alpha_1 = r_1\frac{2\pi}{n_2}+r_2\frac{2\pi}{n_1}=\frac{2\pi}{n_1n_2} \;...(2)$$ Hence, if the problem is soluble for $n = n_1$, and $n = n_2$, it is soluble for $n = n_1n_2$. I can't make sense of it. Can't I always construct a regular $n$ sided polygon for any $n \ge 3 $ ? Also how can it be concluded from $(1)$ and $(2)$ that if problem is solvable for $n=n_1$ and $n=n_2$ it is solvable for $n=n_1n_2$ ?

First question: although it is not mentioned in the excerpt you quoted, Hardy and Wright are referring to constructions using only compass and straight edge (unmarked ruler) in the manner prescribed in geometry at the time of the Ancient Greeks.

Under these restrictions it is impossible, for example, to construct a regular $7$-gon. If you look up "ruler and compass constructions" you will find many online references to this.

Second question: constructing a regular $n$-gon is equivalent to constructing an angle of $2\pi/n$. If you can do this, it is easy to construct twice this angle, three times and so on. Further, if you can construct angles $\alpha$ and $\beta$ then it is easy to construct angles $\alpha\pm\beta$. So, if you can construct angles of $2\pi/n_1$ and $2\pi/n_2$, then you can construct $$r_1\frac{2\pi}{n_2}+r_2\frac{2\pi}{n_1}\ ,$$ and as shown by the above equation, this is the angle $2\pi/(n_1n_2)$.